Question

If \( \alpha = {^nC_r} \) and \( \beta = {^nC_{r-1}} \), then \( 1 + \frac{\alpha}{\beta} \) is equal to

• A. \( \frac{n+1}{r-1} \)
• B. \( \frac{n+1}{r} \)
• C. \( \frac{n-1}{1} \)
• D. \( \frac{n-r+1}{r} \)
• E. \( \frac{n+1}{r+1} \)

Hint:

Memorize ratio: \( \frac{{^nC_r}}{{^nC_{r-1}}} = \frac{n-r+1}{r} \) — very useful shortcut.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem involves the ratio of two consecutive binomial coefficients, which is a standard identity.
Step 2: Key Formula or Approach:
The formula for the ratio of consecutive binomial coefficients is: \[ \frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n-r+1}{r} \] We will first calculate the ratio \(\frac{\alpha}{\beta}\) using this formula and then add 1 to the result.
Step 3: Detailed Explanation:
We are given \(\alpha = {}^nC_r\) and \(\beta = {}^nC_{r-1}\). First, let's find the ratio \(\frac{\alpha}{\beta}\): \[ \frac{\alpha}{\beta} = \frac{{}^nC_r}{{}^nC_{r-1}} \] Let's derive the formula for the ratio: \[ \frac{{}^nC_r}{{}^nC_{r-1}} = \frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r-1)!(n-(r-1))!}} = \frac{n!}{r!(n-r)!} \cdot \frac{(r-1)!(n-r+1)!}{n!} \] \[ = \frac{(r-1)! \cdot (n-r+1)(n-r)!}{r(r-1)! \cdot (n-r)!} = \frac{n-r+1}{r} \] Now we need to calculate \(1 + \frac{\alpha}{\beta}\): \[ 1 + \frac{n-r+1}{r} \] To add these terms, find a common denominator, which is `r`: \[ \frac{r}{r} + \frac{n-r+1}{r} = \frac{r + (n-r+1)}{r} \] \[ = \frac{r + n - r + 1}{r} = \frac{n+1}{r} \] Step 4: Final Answer:
The value of \(1 + \frac{\alpha}{\beta}\) is \(\frac{n+1}{r}\).